What is the solution for x^{''}+4x^'+5x=0,x(0)=0,x^'(0)=4 ?

The solution for x^{''}+4x^'+5x=0,x(0)=0,x^'(0)=4 is x=4e^{-2t}sin(t)

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x^{''}+4x^'+5x=0,x(0)=0,x^'(0)=4

x^{^{''}} + 4 x^{^{'}} + 5 x = 0, x (0) = 0, x^{^{'}} (0) = 4

x = 4 e^{- 2 t} sin (t)

Solution steps

x^{''} (t) + 4 x^{'} (t) + 5 x = 0

Solve linear ODE:

x = 4 e^{- 2 t} sin (t)

x = 4 e^{- 2 t} sin (t)

y^{^{''}} + 81 y = 0, y (0) = 3, y^{^{'}} (0) = 8

14 y^{^{''}} - 15 y^{^{'}} + y = 0, y (0) = 14, y^{^{'}} (0) = \frac{1}{14}

5 y^{^{''}} - 7 y^{^{'}} - 3 y = 0

y^{^{''}} - 6 y^{^{'}} + 18 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{1} (0) = x^{e} - x

What is the solution for x^{''}+4x^'+5x=0,x(0)=0,x^'(0)=4 ?
The solution for x^{''}+4x^'+5x=0,x(0)=0,x^'(0)=4 is x=4e^{-2t}sin(t)